A very basic ring theory question, which I am not able to solve. How does one show that

\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}.

Extending the result: \mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}, if a,b are relatively prime.

My attempt was to define a map, \varphi:\mathbb{Z}[i] \to \mathbb{Z}/10\mathbb{Z} and show that the kernel is the ideal generated by \langle{3-i\rangle}. But I couldn’t think of such a map. Anyhow, any ideas would be helpful.

**Answer**

This diagram shows the Gaussian integers modulo 3-i.

The red points shown are all considered to be 0 but their locations in \mathbb Z[i] are 0, 3-i, i(3-i) and 3-i + i(3-i). Every congruence class must be inside that box once and you can see there are 10 of them.

The arrows show adding by 1 each time. Doing that takes you through every equivalence class and then back to the start.

So \mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}.

**Attribution***Source : Link , Question Author : Community , Answer Author : quanta*